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FREE ENERGY AND EQUILIBRIA ΔG ΔG > 0 reverse process is

FREE ENERGY AND EQUILIBRIA

ΔG < 0 process is spontaneous
ΔG > 0 reverse

process is spontaneous
ΔG = 0 no change is spontaneous: Equilibrium
Le Châtelier’s principle gives us some understanding of how the equilibrium of a system changes when we perturb it.
We are interested in how systems tend towards equilibrium and what obstacles keep them out of equilibrium.
The free energy, or more correctly, the chemical potential is our measure of roughly how far we have to go to come to equilibrium. It is a potential energy of sorts.

"Any change in one of the variables that determines the state of a system in equilibrium causes a shift in the position of equilibrium in a direction that tends to counteract the change in the variable under consideration."

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CHEMICAL POTENTIAL The chemical potential can be used to give quantitative

CHEMICAL POTENTIAL

The chemical potential can be used to give quantitative meaning

to Le Châtelier’s principle.
Chemical potential of component A, µA, is defined as the partial molar Gibbs free energy:
This is the change in G with respect to a infinitesimal change in the amount of component A with all other parameters held constant.
It is essentially the free energy increase (or decrease) associated with adding a little of A to the system.
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DIRECTIONALITY OF A CHEMICAL REACTION Consider a closed system of four

DIRECTIONALITY OF A CHEMICAL REACTION

Consider a closed system of four components

(A, B, C, and D) undergoing a reversible chemical reaction:
In a closed system,
At constant T and P,
At equilibrium:
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EXAMPLE H2O (l) H2O (g) 100 °C This is at constant

EXAMPLE

H2O (l) H2O (g)

100 °C

This is at constant T and P. Its

reversible. So,
ΔG°vap= 0 = ΔH°vap-T ΔS°vap
Note that this implies at equilibrium:
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CHEMICAL POTENTIAL AND PARTIAL PRESSURE We found last time that: (T

CHEMICAL POTENTIAL AND PARTIAL PRESSURE

We found last time that: (T =

constant)
Defining the standard state:
Then,
and,
so,
In a mixture, the chemical potential of A can thus be expressed in terms of its partial pressure and its standard (pure) chemical potential:
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ΔG OF MIXING Consider the isobaric, isothermal mixing of two gases:

ΔG OF MIXING

Consider the isobaric, isothermal mixing of two gases:

Gas A
at


P Atm

Gas B
at
P Atm

Gas A+B
at
P Atm (total)

Mole Fraction component i

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EQUILIBRIUM CONSTANT The Haber process of nitrogen fixation: N2 + 3H2

EQUILIBRIUM CONSTANT

The Haber process of nitrogen fixation: N2 + 3H2 2NH3

Rewrite

letting unitless pressure ratio. No units inside “ln”.

We have used the identity:

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EQUILIBRIUM CONSTANT At any temperature and pressure there exists and equilibrium

EQUILIBRIUM CONSTANT

At any temperature and pressure there exists and equilibrium state

for this reaction.
A combination of such that ΔG = 0.

This equation holds for equilibrium values of

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EQUILIBRIUM CONSTANT Define equilibrium constant (at constant T and P): More

EQUILIBRIUM CONSTANT

Define equilibrium constant (at constant T and P):

More generally: aA +

bB cD + dD

Most generally:

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CHEMICAL POTENTIAL EXAMPLE Reactants Products Reactants Products Δμ° Δμ° μ° Chemical

CHEMICAL POTENTIAL EXAMPLE

Reactants

Products

Reactants

Products

Δμ°

Δμ°

μ°

Chemical potential is a measure of the thermodynamic free

energy.
It tells us what the equilibrium distribution of reactants and products must be
It does not tell us the kinetic rate.

Activation barrier

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EQUILIBRIUM EXAMPLE Oxidation of CO: 2 CO (g) + O2 (g)

EQUILIBRIUM EXAMPLE

Oxidation of CO: 2 CO (g) + O2 (g) 2 CO2

(g)
The free energy change for this reaction is simply:
ΔG ° rxn= 2 ΔG°289(CO2) - 2 ΔG°289(CO) - ΔG°289(O2)
We can calculate this using numbers from the appendix:
ΔG ° rxn= 2 (-394.36) - 2( -137.17) - 0 (kJ/mol) = -514.38 kJ/mol
A relatively large negative number.
ΔG ° rxn= -RTlnK K = e514,380/RT= e207 =
The equilibrium for this reaction lies far in favor of the products.
Large negative ΔG means the reaction goes forward with high probability.
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EQUILIBRIUM EXAMPLE Reactants (CO,O2) Products (CO2) Δμ°, ΔG° The conversion of

EQUILIBRIUM EXAMPLE

Reactants
(CO,O2)

Products (CO2)

Δμ°, ΔG°

The conversion of CO and O2 to CO2

is
energetically favorable, but the reaction is slow
1) We have to first break an O-O double bond
2) The resultant atoms of oxygen must
then react with CO.

For other reactions like
2 H2(g) + O2(g) 2 H2O (l)
We need a catalyst (e.g. platinum surface), and ideally we wish to convert the released energy to work.
Catalysts enable a reaction to take an alternative path, with lower activation barrier.

Reaction coordinate

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ΔH0 FOR SOME NONCOVALENT INTERACTIONS Na+(g) + Cl-(g) NaCl(s) Ionic -785

ΔH0 FOR SOME NONCOVALENT INTERACTIONS

Na+(g) + Cl-(g) NaCl(s) Ionic -785
NaCl(s) Na+(aq) +

Cl-(aq) Ionic +ion-dipole +4
Ar(g) Ar(s) London (fluctuation dipole) -8
Acetone(g) Acetone(l) London-van der Waals -30
H3C H3C (permanent dipole)
2 O O H O Hydrogen bond -20
H H CH3
H
H2NON H
H OH2 H2NON Hydrogen bond (aq) -5
H2NONH2 H O(NH2)2
HOH (Urea (aq))
C3H6(l) + H2O(l) C3H6(aq) Hydrophobic -10
(TSWP p. 98)

Reaction Interaction ΔH0(kJ mol-1)

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PROTEIN UNFOLDING Proteins have a native state. (Really, they tend to

PROTEIN UNFOLDING

Proteins have a native state. (Really, they tend to have

a tight cluster of native states.)
Denaturation occurs when heat or denaturants such as guanidine, urea or detergent are added to solution. Also, the pH can affect folding.
When performing a denaturation process non-covalent interactions are broken.
Ionic, van der-Waals, dipolar, hydrogen bonding, etc.
Solvent is reorganized.
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PROTEIN UNFOLDING Let’s consider denaturation with heat. We can determine a

PROTEIN UNFOLDING

Let’s consider denaturation with heat. We can determine a great

deal about the nature of the protein from such a consideration.
The experimental technique we use for measuring thermodynamic changes here is the differential scanning calorimeter.
Basic experiment: Add heat to sample, measure its temperature change.

heat

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PROTEIN UNFOLDING In differential scanning calorimetry you have two samples: Your

PROTEIN UNFOLDING

In differential scanning calorimetry you have two samples:
Your material

of interest
Control
You put in an amount of heat to raise the temperature of the control at a constant rate, then measure the rate of change in temperature of the other sample as a function of the input heat.
This is a measure of the heat capacity!

Protein
+
Solvent

Solvent

Heat

T1

T2

T1-T2

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PROTEIN UNFOLDING Data for glyceraldehyde-3-phosphate dehydrogenase. pH8.0 pH6.0 Bacillus stearothermophilus E.

PROTEIN UNFOLDING

Data for glyceraldehyde-3-phosphate dehydrogenase.

pH8.0

pH6.0

Bacillus stearothermophilus

E. coli

Rabbit

Is the protein more stable

at pH 8 or 6? Why is B. stear. more stable?